A) \[\int{\frac{\cos 2x-1}{\cos 2x+1}}.\mathbf{dx}\]
B) \[\log \left( {{e}^{x}}+\sqrt{{{e}^{2x}}-1} \right)-{{\sec }^{-1}}({{e}^{x}})+c\]
C) \[\log \left( {{e}^{x}}-\sqrt{{{e}^{2x}}-1} \right)-{{\sec }^{-1}}({{e}^{x}})+c\]
D) None of these
Correct Answer: B
Solution :
[b] \[I=\int{\sqrt{\frac{{{e}^{x}}-1}{{{e}^{x}}+1}}}.dx\] Let \[{{e}^{x}}=sec\theta \] Now, \[{{e}^{x}}.dx=sec\theta .\tan \theta .d\theta \] Now, \[I=\int{\sqrt{\frac{\sec \theta -1}{\sec \theta +1}}}\times \tan \theta .d\theta =\int{\sqrt{\frac{1-\cos \theta }{1+\cos \theta }}}.\tan \theta .d\theta \] \[=\int{\sqrt{\frac{1-\cos \theta }{1+\cos \theta }\times \frac{1-\cos \theta }{1-\cos \theta }}}.\tan \theta .d\theta =\int{\frac{1-\cos \theta }{sin\theta }}\times \frac{sin\theta }{\cos \theta }.d\theta \] \[=\int{\frac{1-\cos \theta }{\cos \theta }}.d\theta =\int{(\sec \theta -1).d\theta }\] \[=log(sec\theta +tan\theta )-\theta +c=log\left( {{e}^{x}}+\sqrt{{{e}^{2x}}-1} \right)-se{{c}^{-1}}({{e}^{x}})+c.\] Hence, option [b] is correct.
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